Metamath Proof Explorer


Theorem nfrexd

Description: Deduction version of nfrex . Usage of this theorem is discouraged because it depends on ax-13 . See nfrexdw for a version with a disjoint variable condition, but not requiring ax-13 . (Contributed by Mario Carneiro, 14-Oct-2016) (New usage is discouraged.)

Ref Expression
Hypotheses nfrald.1 ⊢ Ⅎ 𝑦 𝜑
nfrald.2 ⊢ ( 𝜑 → Ⅎ 𝑥 𝐴 )
nfrald.3 ⊢ ( 𝜑 → Ⅎ 𝑥 𝜓 )
Assertion nfrexd ( 𝜑 → Ⅎ 𝑥 ∃ 𝑦 ∈ 𝐴 𝜓 )

Proof

Step Hyp Ref Expression
1 nfrald.1 ⊢ Ⅎ 𝑦 𝜑
2 nfrald.2 ⊢ ( 𝜑 → Ⅎ 𝑥 𝐴 )
3 nfrald.3 ⊢ ( 𝜑 → Ⅎ 𝑥 𝜓 )
4 dfrex2 ⊢ ( ∃ 𝑦 ∈ 𝐴 𝜓 ↔ ¬ ∀ 𝑦 ∈ 𝐴 ¬ 𝜓 )
5 3 nfnd ⊢ ( 𝜑 → Ⅎ 𝑥 ¬ 𝜓 )
6 1 2 5 nfrald ⊢ ( 𝜑 → Ⅎ 𝑥 ∀ 𝑦 ∈ 𝐴 ¬ 𝜓 )
7 6 nfnd ⊢ ( 𝜑 → Ⅎ 𝑥 ¬ ∀ 𝑦 ∈ 𝐴 ¬ 𝜓 )
8 4 7 nfxfrd ⊢ ( 𝜑 → Ⅎ 𝑥 ∃ 𝑦 ∈ 𝐴 𝜓 )